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zubka84 [21]
3 years ago
7

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

sphere is 5 cm and that of the larger sphere is 12 cm. The electric field at the surface of the larger sphere is 660 kV/m. Find the surface charge density on each sphere.
Physics
1 answer:
mrs_skeptik [129]3 years ago
6 0

Answer:

surface charge density on each sphere is 440 \times 10^{-9} C

Explanation:

given data

radius of smaller sphere = 5 cm

radius of  larger sphere is 12 cm

electric field at surface of larger sphere = 660 kV/m = 660 × 1000 v/m

solution

we apply here electric field formula that is express as

E = (\frac{1}{4\pi\epsilon  })\times  (\frac{Q_{1} }{R^{2} } )    .................1

put here value

660000 = 9 \times 10^9 \times \frac{Q1}{0.12^2}  

Q1 = 1056 × 10^{-9}

and

here field inside a conductor is zero so that electric potential ( V ) is constant

\frac{Q{1} }{R} = \frac{Q{2} }{r}   ..................2

so Q2 will be

Q2 =  \frac{5}{12} \times 1056 \times 10^{-9}  

Q2 =  440 \times 10^{-9}  C

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The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance  between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle \gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120 is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which  is also the net displacement):

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c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

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